SOLUTION: A certain radioactive isotope has a​ half-life of approximately 78 years. How many years would be required for a given amount of this isotope to decay to 60​% of that​ amount

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Question 1162420: A certain radioactive isotope has a​ half-life of approximately 78 years. How many years would be required for a given amount of this isotope to decay to 60​% of that​ amount? ​(​Hint: first find k. Then use k to find​ t)

Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39618)   (Show Source): You can put this solution on YOUR website!
MODEL Form:


If half-life 78 years, then .
.


When any given amount to go to 60% of the starting amount?
------------solve this for x.

Answer by ikleyn(52792)   (Show Source): You can put this solution on YOUR website!
.

We are given that the half-life period is 78 years;  therefore, we can write


    p(t) = ,      (1)


where p(t) is the remaining mass fraction. (It is the standard radioactive decay model in terms of half-life period).


The problem asks to determine the time "t" when p(t) = 0.6.


In this case, the equation (1) takes the form


    0.6 = .


Take logarithm base 2 of both sides


     = 


    t =  =  = 57.48 years.    ANSWER

Solved.

---------------

See the lesson
    - Radioactive decay problems
in this site and find there many other similar solved problems.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Logarithms".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.


/\/\/\/\/\/\/\/

The lesson to learn from my post is THIS:

    If you are given input data in terms of half-life, you do not need to convert your data 

    into ekt-model.  Such conversion is an excessive work and unnecessary calculations.


    You can complete all calculations in terms of the half-life model, working consistently with degrees of 2, 
    which is your base in this case..



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